sum over reciprocal of primes times coefficient I would like to show that
$$
\sum_{p\leq x} \frac{1}{p^{1+2/\log x}}\left(\frac{\log\left(x/p\right)}{\log(x)}\right)^2=\log\log x +\mathcal{O}(1)
$$

What I have tried
Since we know that
$$
\sum_{p\leq x}\frac{1}{p}=\log\log x+\mathcal{O}(1)
$$
(see this post) I thought I could use Abel's summation to prove the above estimate. Abel's summation says that given a sequence of real numbers $(a_n)_{n=0}^{\infty}$, we can define the partial sum
$$
A(t)=\sum_{n\leq t}a_n
$$
and, if $\phi$ is a continously differentiable function on $[1,x]$, we have
$$
\sum_{1\leq n\leq x}a_n\phi(n)=A(x)\phi(x)-\int_1^x A(u)\phi^{\prime}(u)\,du
$$
In our case I would take
$$
a_n = \begin{cases} 1/p &\text{if } n=p \text{ is prime}, \\
 0 &\text{otherwise}, \end{cases} 
$$
and
$$
\phi(t)=\frac{1}{t^{2/\log x}}\left(\frac{\log \left(x/t\right)}{\log x}\right)^2
$$
but in this way I have
$$
\phi(x)=0
$$
and therefore I am only left with the integral from Abel's summation formula which looks pretty ugly to me
$$
\sum_{p\leq x} \frac{1}{p^{1+2/\log x}}\left(\frac{\log\left(x/p\right)}{\log(x)}\right)^2=\int_1^x \left(\sum_{p\leq u}\frac{1}{p}\right)\frac{2\log\left(x/u\right)\left(\log(x/u)+\log x\right)}{u^{2/\log x+1}(\log x)^3}\,du
$$
Is there any another way I could proceed? Do you have any hint? 
Thank you for your help!

P.S.
Since 
$$
\left(\frac{\log\left(x/p\right)}{\log(x)}\right)^2\leq 1
$$
and
$$
\frac{1}{p^{2/\log x}}\leq 1
$$
I clearly have that
$$
\sum_{p\leq x} \frac{1}{p^{1+2/\log x}}\left(\frac{\log\left(x/p\right)}{\log(x)}\right)^2\leq \sum_{p\leq x}\frac{1}{p}=\log\log x+\mathcal{O}(1)
$$
hence I only need to prove that I similar lower bound holds too.
 A: Here is a general strategy. You are interested in
$$\sum_{p \leq x} \frac{1}{p^{1 + \frac{2}{\log x}}} - \frac{2}{\log x} \sum_{p \leq x} \frac{\log p}{p^{1 + \frac{2}{\log x}}} + \frac{1}{(\log x)^2} \sum_{p \leq x} \frac{(\log p)^2}{p^{1 + \frac{2}{\log x}}}.$$
We use partial summation, noting that
$$\sum_{p \leq x} \frac{1}{p} = \log \log x + b + O\left(\frac{1}{\log x}\right), \qquad \sum_{p \leq x} \frac{\log p}{p} = \log x + O(1), \qquad \sum_{p \leq x} \frac{(\log p)^2}{p} = \frac{(\log x)^2}{2} + O(1).$$
Here $b$ is some explicit constant.
The first terms via partial summation are
$$e^{-2} \log \log x + b e^{-2}, \qquad -2e^{-2}, \qquad \frac{e^{-2}}{2}.$$
The error is $O(1/\log x)$ for the first two and $O(1/(\log x)^2)$ for the third.
For the second terms, we note that
$$\frac{d}{dt} \frac{1}{t^{\frac{2}{\log x}}} = -\frac{2}{\log x} \frac{1}{t} \exp\left(-\frac{2 \log t}{\log x}\right),$$
and so we must evaluate
$$\frac{2}{\log x} \int_{2}^{x} \log \log t \exp\left(-\frac{2\log t}{\log x}\right) \, \frac{dt}{t}, \qquad \frac{2b}{\log x} \int_{2}^{x} \exp\left(-\frac{2\log t}{\log x}\right) \, \frac{dt}{t}, \qquad -\frac{4}{(\log x)^2} \int_{2}^{x} \log t \exp\left(-\frac{2\log t}{\log x}\right) \, \frac{dt}{t}, \qquad \frac{1}{(\log x)^3} \int_{2}^{x} (\log t)^2 \exp\left(-\frac{2\log t}{\log x}\right) \, \frac{dt}{t}.$$
(There are also error terms; the method below shows that they are $O(\log \log x/\log x)$.) We make the change of variabes $t \mapsto e^t$, so that these become
$$\frac{2}{\log x} \int_{\log 2}^{\log x} \log t \exp\left(-\frac{2 t}{\log x}\right) \, dt, \qquad \frac{2b}{\log x} \int_{\log 2}^{\log x} \exp\left(-\frac{2 t}{\log x}\right) \, dt, \qquad -\frac{4}{(\log x)^2} \int_{\log 2}^{\log x} t \exp\left(-\frac{2 t}{\log x}\right) \, dt, \qquad \frac{1}{(\log x)^3} \int_{\log 2}^{\log x} t^2 \exp\left(-\frac{2 t}{\log x}\right) \, dt.$$
The latter three can be explicitly evaluated (directly for the second, via integration by parts for the last two); they give
$$-b e^{-2} - b \exp\left(-\frac{2 \log 2}{\log x}\right), \qquad 3e^{-2} - \exp\left(-\frac{2 \log 2}{\log x}\right) \left(\frac{2 \log 2}{\log x} + 1\right), \qquad -\frac{5}{4} e^{-2} + \frac{1}{4} \exp\left(-\frac{2 \log 2}{\log x}\right) \left(\frac{2 (\log 2)^2}{(\log x)^2} + \frac{2 \log 2}{\log x} + 1\right).$$
Note that $\exp(-2\log 2/\log x) = 1 + O(1/\log x)$. For the former, we integrate by parts once, getting
$$-e^{-2} \log \log x + \log \log 2 \exp\left(-\frac{2 \log 2}{\log x}\right) + \int_{\log 2}^{\log x} \frac{1}{t} \exp\left(-\frac{2 t}{\log x}\right) \, dt.$$
For this second term, we make the change of variables $t \mapsto \frac{\log x}{2} t$, so that this becomes
$$\int_{\frac{2 \log 2}{\log x}}^{2} \frac{e^{-t}}{t} \, dt = E_1\left(\frac{2 \log 2}{\log x}\right) - E_1(2),$$
where $E_1(z)$ is the exponential integral. Since $E_1(z) = -\log z - \gamma_0 + O(z)$ about $z = 0$, we get additional terms
$$\log \log x - \log 2 - \log \log 2 - \gamma_0 - E_1(2) + O\left(\frac{1}{\log x}\right).$$
So now we combine everything and find that the desired asymptotic is
$$\log \log x - b + \frac{e^{-2}}{4} + \frac{1}{4} - \log 2 - \gamma_0 - E_1(2) + O\left(\frac{\log \log x}{\log x}\right).$$

Alternatively, you can use your approach to get the expression
$$\frac{4}{\log x} \int_{2}^{x} \log \log t \exp\left(-\frac{2 \log t}{\log x}\right) \, \frac{dt}{t} - \frac{6}{(\log x)^2} \int_{2}^{x} \log t \log \log t \exp\left(-\frac{2 \log t}{\log x}\right) \, \frac{dt}{t} + \frac{2}{(\log x)^3} \int_{2}^{x} (\log t)^2 \log \log t \exp\left(-\frac{2 \log t}{\log x}\right) \, \frac{dt}{t}.$$
(Note that you seem to have calculated the derivative incorrectly.) We again make the change of variables $t \mapsto e^t$, yielding
$$\frac{4}{\log x} \int_{\log 2}^{\log x} \log t \exp\left(-\frac{2 t}{\log x}\right) \, dt - \frac{6}{(\log x)^2} \int_{\log 2}^{\log x} t \log t \exp\left(-\frac{2 t}{\log x}\right) \, dt + \frac{2}{(\log x)^3} \int_{\log 2}^{\log x} t^2 \log t \exp\left(-\frac{2 t}{\log x}\right) \, dt.$$
Now integrate by parts repeatedly, integrating the exponential and differentiating everything else.
